d******g
发帖数: 34 | 1 the interviewer is a quant from prime brokerage. A very nice guy.
1. Roll a penny around another fixed penny in the center with edges in close
contact. After moving half circle around the center penny, you will find
the penny in motion has rotated 360 deg. Why?
(found this question in an older post)
2. find a way to express any rational number n = product(p1!, p2!, …) /
product(q1!, q2!, …), here p1, p2… and q1, q2.. are all prime numbers and
! means factorial.
thank you. | a******n
发帖数: 11246 | 2 2. 先对n=自然数归纳。如果n+1是合数，那么分解一下，每个因子都能表示成那个形式
如果n+1质数，n+1=(n+1)! / n!，分母n!中每个部分都能表示成那个形式。
然后对一般n=t/s, t和s都能以这个形式表达，所以t/s也可以。
close
and
【在 d******g 的大作中提到】
: the interviewer is a quant from prime brokerage. A very nice guy.
: 1. Roll a penny around another fixed penny in the center with edges in close
: contact. After moving half circle around the center penny, you will find
: the penny in motion has rotated 360 deg. Why?
: (found this question in an older post)
: 2. find a way to express any rational number n = product(p1!, p2!, …) /
: product(q1!, q2!, …), here p1, p2… and q1, q2.. are all prime numbers and
: ! means factorial.
: thank you.
| R******t
发帖数: 2648 | 3 1.penny in motion相对fixed penny转了360度
2.any ration number r=p/q=(p!(q-1)!)/(q!(p-1)!)
if q-1 and p-1 are both prime numbers, done
if either is not, WLOG, p-1 is not prime number, write p-1=p_1p_2...p_m,
where p_1,...,pm are prime numbers, then
p-1=(p_1!p_2!...p_m!)/((p_1-1)!...(p_m-1)!)
repeat until all numbers are prime ones. Cancel same terms in numerator and denominator and discard 1! terms.
close
and
【在 d******g 的大作中提到】
: the interviewer is a quant from prime brokerage. A very nice guy.
: 1. Roll a penny around another fixed penny in the center with edges in close
: contact. After moving half circle around the center penny, you will find
: the penny in motion has rotated 360 deg. Why?
: (found this question in an older post)
: 2. find a way to express any rational number n = product(p1!, p2!, …) /
: product(q1!, q2!, …), here p1, p2… and q1, q2.. are all prime numbers and
: ! means factorial.
: thank you.
| a******n
发帖数: 11246 | 4 如果q-1是质数，q就不是了，呵呵
and
find
【在 R******t 的大作中提到】
: 1.penny in motion相对fixed penny转了360度
: 2.any ration number r=p/q=(p!(q-1)!)/(q!(p-1)!)
: if q-1 and p-1 are both prime numbers, done
: if either is not, WLOG, p-1 is not prime number, write p-1=p_1p_2...p_m,
: where p_1,...,pm are prime numbers, then
: p-1=(p_1!p_2!...p_m!)/((p_1-1)!...(p_m-1)!)
: repeat until all numbers are prime ones. Cancel same terms in numerator and denominator and discard 1! terms.
:
: close
: and
| R******t
发帖数: 2648 | 5 2，3
【在 a******n 的大作中提到】
: 如果q-1是质数，q就不是了，呵呵
:
: and
: find
| c******r
发帖数: 300 | 6 It doesn't matter since (q-1)! can be factorized as products of prime
numbers that are less than q.
【在 a******n 的大作中提到】
: 如果q-1是质数，q就不是了，呵呵
:
: and
: find
| p******5
发帖数: 138 | 7 This is essential!
Actually it is sufficient to check integers only.
1,2,3 are all true.
Assume m <= n are all true. Let's consider about n+1.
n+1 = (n+1)!/n!, if n+1 is prime, then n! can be decomposed by products of
prime numbers which are less than n. Use the assumption to get conclusion.
If n+1 isn't prime, (n+1)! still can be decomposed by products of
prime numbers which are less than n.
【在 c******r 的大作中提到】
: It doesn't matter since (q-1)! can be factorized as products of prime
: numbers that are less than q.
| G********d
发帖数: 10250 | 8 How do you define rational numbers?
If you define it as fraction field of the integer ring \mathbb{Z},
then this question comes naturally from the fact that \mathbb{Z} is a UFD. | w*********m
发帖数: 196 | 9 (p-1)!=(p_1!p_2!...p_m!)/((p_1-1)!...(p_m-1)!)????
应该是(p-1)=p_1p_2...p_m=(p_1!p_2!...p_m!)/((p_1-1)!...(p_m-1)!)吧 | w*********m
发帖数: 196 | 10 赞这个答案
of
conclusion.
【在 p******5 的大作中提到】
: This is essential!
: Actually it is sufficient to check integers only.
: 1,2,3 are all true.
: Assume m <= n are all true. Let's consider about n+1.
: n+1 = (n+1)!/n!, if n+1 is prime, then n! can be decomposed by products of
: prime numbers which are less than n. Use the assumption to get conclusion.
: If n+1 isn't prime, (n+1)! still can be decomposed by products of
: prime numbers which are less than n.
| w*********m
发帖数: 196 | 11 第一题有证明么？
我有一个解法，如果两个硬币一个以顺时针绕中心旋转，另外一个以逆时针绕中心旋转
。如果两个硬币的角速度的大小时时刻刻相等，那么在这个坐标系里按照对称性，接触
点的位置不变。所以一个角速度+1，另外一个-1.
假设观察者站在其中一个硬币的边缘，那么在他的旋转坐标系里他看到的情况就是题中
描述的样子，而且另外的那个硬币相对于观察者的坐标系旋转速度是1-(-1)=2. | k*******a
发帖数: 772 | 12 第一题高中物理题啊，第一个penny的圆心移动了 pi*2R=2pi*R (因为围着第一个penny
做2R为半径的旋转）
但是相对第一个penny的圆心，接触点是做R的圆周运动，所以转了2pi | w*********m
发帖数: 196 | 13 必须用一个条件，接触点在静止的penny里的瞬时速度是0 (no slippery)，而移动的
penny圆心的速度是w*2R (w 是移动penny圆心相对于静止penny运动的角速度)，所以接
触点相对于移动的penny的坐标系的顺时速度是－w*2R，而接触点相对于移动的penny的
角速度是-w*2R/R等于-2w. |
|
